Question

If $53^{53}-33^3$ is divided by $10,$ then the remainder obtained is

Answer 1

Answer: 6

We know that,
$x^n-y^n=\left(x-y\right)\left(x^{n-1}+x^{n-2}y+....+x{y}^{n-2}+y^{n-1}\right)$
i.e. $x^n-y^n$ is divisible by $x-y$
$53^{53}-33^3=53^{53}-3^{53}+3^{53}-33^3+3^3-3^3$
$=\left(53^{53}-3^{53}\right)+\left(3^{53}-3^3\right)-\left(33^3-3^3\right)$
Clearly, $1^{st}$ bracket is divisible by $50$ and the last one is divisible by $30.$ So, both are divisible by $10.$
$3^{53}-3^3=3^3\cdot 3^{50}-3^3=3^3\left(3^{50}-1\right)$
$=3^3\left({\left(10-1\right)}^{25}-1\right)=27{\left(10-1\right)}^{25}-27=27\left(10k-1\right)-27$
$=270k-54=270k-60+6=10\mu +6.$
Hence, the remainder obtained is $6.$